1. To compute a student's Grade Point Average (GPA) for a term,
the student's grades for each course are weighted by the number of
credits for the course. Suppose a student had these grades:
4.0 in a 5 credit Math course
2.1 in a 2 credit Music course
2.8 in a 5 credit Chemistry course
3.0 in a 4 credit Journalism course
What is the student's GPA for that term? Round to two decimal
places. Student's GPA =
2.A set of exam scores is normally distributed with a mean = 80
and standard deviation = 6.
Use the Empirical Rule to complete the following
sentences.
68% of the scores are between ____ and _____ .
95% of the scores are between _____ and _____ .
99.7% of the scores are between ______ and _______
3. To compute a student's Grade Point Average (GPA) for a term,
the student's grades for each course are weighted by the number of
credits for the course. Suppose a student had these grades:
3.7 in a 5 credit Math course
2.4 in a 2 credit Music course
2.9 in a 4 credit Chemistry course
2.8 in a 4 credit Journalism course
What is the student's GPA for that term? Round to two decimal
places.
Student's GPA =
In: Statistics and Probability
Your marketing department believes it has a new commercial that
will increase the percent of people who plan on purchasing from one
of your stores in the next month. You take a sample of people and
before you show them the new commercial, ask them if they are
planning on purchasing from one of your stores in the next month.
You then show them the new commercial and follow up by asking again
if they plan on purchasing from one of your stores in the next
month. In your data analysis, you look at the difference of
(post-commercial)-(pre-commercial). You want to test the claim that
there is no difference between the pre and post commercial mean
percentages, and do so at the α=0.02α=0.02 level.
You believe the population of difference scores is normally
distributed, but you do not know the standard deviation. You obtain
the following sample of data:
pre-commercial | post-commercial |
---|---|
55.6 | 54.1 |
53.3 | 40.3 |
75.6 | 67 |
49 | 81.1 |
52.3 | 79 |
58.8 | -44.7 |
67.7 | 78.7 |
63.5 | -47 |
55.8 | 24.2 |
58.2 | 27.7 |
52.3 | 81.6 |
63.5 | 74.5 |
38.4 | 4.8 |
45.8 | 25.4 |
50.5 | 34.3 |
45.8 | -21.9 |
48.7 | 7.8 |
51.3 | 9.3 |
65.6 | 94.9 |
62.9 | 37.4 |
63.7 | 24.9 |
47.2 | 26.8 |
60.7 | 23 |
68.8 | 93.1 |
56 | 52 |
In: Statistics and Probability
4. Consider the following time series:
Quarter | Year 1 | Year 2 | Year 3 |
1 | 80 | 74 | 65 |
2 | 69 | 61 | 51 |
3 | 48 | 50 | 43 |
4 | 68 | 71 | 82 |
a. Construct a time-series plot. What type of pattern exists in the data? Is there an indication of a seasonal pattern? (10 points)
b. Use multiple linear regression model with dummy variables as follows to develop an equation to account for seasonal effects in the data. Qtr1 = 1 if quarter 1, 0 else; Qtr2 = 1 if quarter 2, 0 else; Qtr3 = 1 if quarter 3, 0 else. (20 points)
c. Compute the quarterly forecasts for next year. (10 points)
In: Statistics and Probability
The average worker at Acme Labs produces 24 widgets per hour, with a standard deviation of 8 widgets. You have invented a new production process, and you hypothesize that your new process will help workers produce more widgets per hour. After implementing the new production process, you take a sample of workers and measure how many widgets they produce per hour.
a) (2 points) What are the null and alternative hypotheses, in symbols?
b) (2 points) What are the null and alternative hypotheses, in words?
c) (5 points) The average number of widgets produced per hour by a sample of 80 workers turns out to be 26. Calculate the test statistic for this hypothesis test and find the p-value.
d) (4 points) At an alpha level of .05, do we have evidence that the new production process really increases productivity? Explain your reasoning.
In: Statistics and Probability
Find the Banzhaf power distribution of the following weighted voting system:
[q : 15,13,8,6]
a. If q=27
b. If q=22
c. If q=32
In: Statistics and Probability
I need answers for question 3 and 4. I believe I'm correct with question 1 and 2 but not sure which could make question 3 and 4 incorrect.
Coin |
7 |
6 |
4 |
6 |
4 |
6 |
4 |
6 |
4 |
4 |
3 |
5 |
5 |
6 |
3 |
3 |
5 |
6 |
3 |
7 |
5 |
2 |
6 |
6 |
6 |
3 |
3 |
7 |
2 |
4 |
6 |
5 |
6 |
6 |
7 |
Mean: 4.885 Standard deviation: 1.490 |
P(x=0) |
P(x=6) |
|||
P(x=1) |
P(x=7) |
|||
P(x=2) |
P(x=8) |
|||
P(x=3) |
P(x=9) |
|||
P(x=4) |
P(x=10) |
|||
P(x=5) |
4. Give the probability for the following based on the calculations in question 3 above, with the probability of a success being ½. (Complete sentence not necessary; round your answers to three decimal places)
P(x≥2) |
P(x<0) |
|||
P(x>2) |
P(x≤5) |
|||
P(5<x ≤7) |
P(x<5 or x≥7) |
In: Statistics and Probability
Range of ankle motion is a contributing factor to falls among the elderly. Suppose a team of researchers is studying how compression hosiery, typical shoes, and medical shoes affect range of ankle motion.
In particular, note the variables Barefoot and Footwear5 (FW5). Barefoot represents a subject's range of ankle motion (in degrees) while barefoot, and Footwear5 (FW5) represents their range of ankle motion (in degrees) while wearing compression hosiery and medical shoes.
Use this data and your preferred software to calculate the equation of the least-squares linear regression line to predict a subject's range of ankle motion while wearing compression hosiery and medical shoes, ?̂ y^ , based on their range of ankle motion while barefoot, ?x . Round your coefficients to two decimal places of precision.
?̂ y^ =
A physical therapist determines that her patient Jan has a range of ankle motion of 7.26°7.26° while barefoot. Predict Jan's range of ankle motion while wearing compression hosiery and medical shoes, ?̂ y^ . Round your answer to two decimal places.
?̂ =y^=
Suppose Jan's actual range of ankle motion while wearing compression hosiery and medical shoes is 9.79°9.79° . Use her predicted range of ankle motion to calculate the residual associated with this value. Round your answer to two decimal places.
residual=
In order to assess the linear regression equation's ability to predict range of ankle motion, the physical therapist reviewed a scatterplot of the researchers' sample data and calculated the correlation, ?=0.53r=0.53 .
Barefoot | FW1 | FW2 | FW3 | FW4 | FW5 |
34.851 | 32.927 | 37.455 | 31.719 | 27.937 | 27.483 |
17.309 | 18.468 | 11.617 | 23.863 | 26.681 | 20.687 |
30.921 | 32.616 | 31.59 | 37.228 | 27.908 | 32.626 |
23.067 | 28.614 | 23.782 | 23.766 | 25.293 | 21.336 |
26.665 | 26.056 | 24.749 | 24.809 | 30.978 | 28.229 |
24.865 | 28.931 | 31.218 | 22.463 | 25.386 | 21.749 |
23.64 | 37.264 | 27.016 | 27.615 | 29.044 | 30.889 |
27.416 | 30.716 | 22.181 | 27.281 | 31.016 | 37.565 |
18.079 | 20.052 | 17.486 | 11.296 | 20.909 | 14.573 |
19.659 | 21.54 | 20.186 | 20.369 | 17.609 | 22.501 |
32.875 | 30.725 | 28.188 | 28.678 | 29.013 | 29.013 |
12.859 | 16.772 | 16.289 | 12.07 | 14.428 | 24.308 |
23.155 | 24.625 | 18.363 | 24.325 | 31.178 | 22.121 |
21.66 | 31.301 | 25.894 | 25.893 | 27.762 | 25.498 |
21.808 | 24.97 | 19.964 | 23.087 | 25.475 | 20.586 |
27.784 | 20.623 | 21.233 | 30.305 | 27.119 | 26.509 |
26.953 | 33.153 | 27.35 | 23.463 | 33.045 | 24.015 |
21.203 | 25.033 | 20.387 | 33.77 | 28.263 | 23.86 |
26.065 | 34.133 | 22.761 | 26.053 | 29.894 | 23.954 |
16.658 | 27.805 | 15.032 | 26.853 | 27.778 | 20.402 |
30.123 | 28.414 | 27.457 | 26.625 | 27.539 | 23.434 |
15.447 | 22.073 | 15.2 | 33.395 | 22.7 | 20.321 |
23.924 | 25.478 | 19.357 | 20.732 | 29.334 | 20.325 |
13.807 | 24.112 | 21.877 | 20.653 | 26.294 | 26.093 |
16.114 | 16.365 | 12.127 | 17.134 | 23.874 | 17.643 |
22.533 | 29.161 | 30.178 | 25.869 | 31.884 | 19.14 |
23.005 | 18.487 | 19.135 | 21.793 | 20.111 | 19.14 |
In: Statistics and Probability
Statistics for Criminology and Criminal Justice
The probability of being acquitted in criminal court in Baltimore, Maryland, is .40. You take a random sample of the past 10 criminal cases where the defendant had a public defender and find that there were seven acquittals and three convictions. What is the probability of observing seven or more acquittals out of 10 cases if the true probability of an acquittal is .40? By using an alpha of .05, test the null hypothesis (that the probability of an acquittal is .40 for defendants with public defenders), against the alternative hypothesis that it is greater than .40.
In: Statistics and Probability
You wish to test the following claim (HaHa) at a significance level of α=0.01α=0.01.
Ho:μ1=μ2Ho:μ1=μ2
Ha:μ1≠μ2Ha:μ1≠μ2
You believe both populations are normally distributed, but you do not know the standard deviations for either. And you have no reason to believe the variances of the two populations are equal You obtain a sample of size n1=14n1=14 with a mean of ¯x1=79.7x¯1=79.7 and a standard deviation of s1=11.8s1=11.8 from the first population. You obtain a sample of size n2=15n2=15 with a mean of ¯x2=83.8x¯2=83.8 and a standard deviation of s2=19.1s2=19.1 from the second population.
Similarly, you now wish to test the following claim (HaHa) at a significance level of α=0.10α=0.10.
Ho:μ1=μ2Ho:μ1=μ2
Ha:μ1≠μ2Ha:μ1≠μ2
You believe both populations are normally distributed, but you do not know the standard deviations for either. And you have no reason to believe the variances of the two populations are equal You obtain a sample of size n1=14n1=14 with a mean of ¯x1=59.9x¯1=59.9 and a standard deviation of s1=15.6s1=15.6 from the first population. You obtain a sample of size n2=10n2=10 with a mean of ¯x2=51.6x¯2=51.6 and a standard deviation of s2=13.4s2=13.4 from the second population.
In: Statistics and Probability
The 2015 American Time Use Survey contains data on how many minutes of sleep per night an average college student gets. According to this survey, the minutes that college students sleep per night are right skewed with a mean of 529.9 minutes and standard deviation 135.6 minutes.
Suppose we take a random sample of size 20. What is the probability that the mean of the 20 times is greater than 538 minutes?
Answer --> It is not appropriate to calculate probabilities in this situation. (explain why, please)
Suppose we take a random sample of size 60. What is the probability
that the mean of the 60 times is less than 479.4 minutes?
A. 0.0020
B. 0.9980
C. 0.3557
D. 0.6643
E. It is not appropriate to calculate probabilities in this situation.
In: Statistics and Probability
If you draw a card with a value of four or less from a standard deck of cards, I will pay you $134. If not, you pay me $37. (Aces are considered the highest card in the deck.)
Step 2 of 2 :
If you played this game 621 times how much would you expect to win or lose? Round your answer to two decimal places. Losses must be entered as negative.
In: Statistics and Probability
The amounts (in ounces) of juice in eight randomly selected juice bottles are:
15.3 15.3 15.7 15.7
15.3 15.9 15.3 15.9
Construct a 98% confidence interval for the mean amount of juice in all such bottles.
In: Statistics and Probability
Assume that adults have IQ scores that are normally distributed with a mean of 97.7 and a standard deviation 21.8
Find the first quartile Upper Q1, which is the IQ score separating the bottom 25% from the top 75%. (Hint: Draw a graph.)
The first quartile is ________. (Type an integer or decimal rounded to one decimal place as needed.)
In: Statistics and Probability
The manager of a cosmetics section of a large department store wants to determine whether newspaper advertising affects sales. She randomly selects 10 items currently in stock that are priced at their usual competitive value, and she records the quantity of each item sold for a one-week period. Then, without changing their price, she places a large ad in the newspaper, advertising the 10 items. Again, she records the quantity sold for a one-week period. Her data are listed below:
Item |
# Sold Before Ad |
# Sold After Ad |
1 |
18 |
24 |
2 |
3 |
7 |
3 |
16 |
19 |
4 |
23 |
23 |
5 |
15 |
5 |
6 |
7 |
11 |
7 |
13 |
12 |
8 |
16 |
12 |
9 |
20 |
25 |
10 |
14 |
20 |
Test whether placement of the ad resulted in a significant increase in sales, with an alpha of .05.
a. State the H0 and H1 hypotheses, in statistical notation, and identify the critical t value(s).
b. Calculate standard error and the observed t-statistic. What is your decision regarding the null hypothesis? Why?
c. State your results in nonstatistical terms.
d. State the type of t-test you used for problems A, B, and C, above, and explain why each test was the appropriate one to choose.
In: Statistics and Probability
A social psychologist claims to know how many casual friends are in the typical person's social network. The psychologist states that the average number of casual friends is 11. You question the validity of this researcher's claim and decide to conduct your own study. You interview a random sample of people and determine for each the number of friends or social acquaintances they see or talk to at least once a year. Your data are below:
16 |
4 |
6 |
11 |
16 |
4 |
8 |
7 |
8 |
3 |
10 |
6 |
9 |
21 |
14 |
12 |
9 |
11 |
Test if your results differ from the social psychologist's claim, with an alpha of .05.
a. State the H0 and H1 hypotheses, in statistical notation, and identify the critical t value(s).
b. Calculate standard error and the observed t-statistic. What is your decision regarding the null hypothesis? Why?
c. State your results in nonstatistical terms.
In: Statistics and Probability